MATH 308 Lecture 25

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Lecture Notes

Convolution Function

Defined

$(f*g)(t)=\int _{0}^{t}f(t-v)\,g(v)\,\mathrm {d} v=\int _{0}^{t}f(u)\,g(t-u)\,\mathrm {d} u$

• It is associative, so $f*g=g*f$
• It is distributive, so $(f*(g(t)+a\,h(t)))(t)=(f*g)(t)+a(f*h)(t)$
• $f*0(t)=0$

Example

$f(t)=\mathrm {e} ^{t}$, and $g(t)=\sin {(3\,t)}$:

$(f*g)(t)=\int _{0}^{t}\mathrm {e} ^{t-v}\,\sin {3v}\,\mathrm {d} v=\int _{0}^{t}\mathrm {e} ^{t}\,\sin {(3(t-v))}\,\mathrm {d} v$

$f(t)=t^{2}$, and $g(t)=1$

$(f*g)(t)=\int _{0}^{t}v^{2}\cdot 1\,\mathrm {d} v={\frac {t^{3}}{3}}$

Theorem 6.6.1

${\mathcal {L}}\left\{f*g\right\}={\mathcal {L}}\left\{f\right\}\,{\mathcal {L}}\left\{g\right\}$

So finding the inverse Laplace transform of ${\frac {1}{(s-7)^{9}\,(s^{2}+4)}}$ is equivalent to finding the convolution ${\frac {(t^{8}\,\mathrm {e} ^{7t})*(\sin {2t})}{8!\cdot 2}}$

Exercises

1. ${\mathcal {L}}^{-1}\left\{{\frac {1}{(s-1)(s^{2}+9)}}\right\}=\mathrm {e} ^{t}*{\frac {\sin {3t}}{3}}$
2. ${\mathcal {L}}\left\{g(t)\right\}=G(s)$, $F(s)=G(s)\,{\frac {1}{s^{2}+1}}={\mathcal {L}}\left\{g(t)*\sin {t}\right\}$

Exercise 3

Find Laplace transforms of

1. $f(t)=\int _{0}^{t}(t-v)^{3}\,\sin {2v}\,\mathrm {d} v=t^{3}*\sin {2t}$, so ${\mathcal {L}}\left\{f(t)\right\}={\frac {6}{s^{4}}}\,{\frac {2}{s^{2}+4}}={\frac {12}{s^{4}(s^{2}+4)}}$
2. $g(t)=\int _{0}^{t}\mathrm {e} ^{v-t}\,\cos {v}\,\mathrm {d} v=\mathrm {e} ^{-t}*\cos {t}$, so ${\mathcal {L}}\left\{g(t)\right\}={\frac {s}{s^{2}+1}}\,{\frac {1}{s+1}}={\frac {s}{(s^{2}+1)(s+1)}}$

Solving Differential Equations

Find the solution to $y''+\omega ^{2}\,y=g(t)$ when $y(0)=0$ and $y'(0)=1$.

Take Laplace transform of both sides:

$s^{2}{\mathcal {L}}\left\{y\right\}-y'(0)-s\,y(0)+\omega ^{2}\,{\mathcal {L}}\left\{y\right\}={\mathcal {L}}\left\{g(t)\right\}$

Solve for ${\mathcal {L}}\left\{y\right\}$:

${\mathcal {L}}\left\{y\right\}={\frac {{\mathcal {L}}\left\{g(t)\right\}}{s^{2}+\omega ^{2}}}+{\frac {1}{s^{2}+\omega ^{2}}}$

So the solution is

$y=g(t)*{\frac {\sin {\omega \,t}}{\omega }}+{\frac {\sin {\omega \,t}}{\omega }}$

---

Find the solution to $y''+3y'+2y=\cos {a\,t}$ when $y(0)=1$ and $y'(0)=0$

We can solve this using two methods: undetermined coefficients and variation of parameters. Let's try to use Laplace transforms:

$s^{2}\,{\mathcal {L}}\left\{y\right\}-y'(0)-s\,y(0)+3{\mathcal {L}}\left\{y\right\}-3y(0)+2{\mathcal {L}}\left\{y\right\}={\frac {s}{s^{2}+a^{2}}}$

Solve for ${\mathcal {L}}\left\{y\right\}$:

${\begin{aligned}{\mathcal {L}}\left\{y\right\}&={\frac {s}{(s^{2}+a^{2})(s^{2}+3s+2)}}+{\frac {s+3}{s^{2}+3s+2}}\\&=\left({\frac {s}{s^{2}+a}}\right)\,\left({\frac {1}{s^{2}+3s+2}}\right)+{\frac {s+3}{(s+2)(s+1)}}\\&={\mathcal {L}}\left\{\cos {a\,t}\right\}\,{\mathcal {L}}\left\{\mathrm {e} ^{-2t}\right\}\,{\mathcal {L}}\left\{\mathrm {e} ^{-t}\right\}+{\frac {s+3}{(s+2)(s+1)}}\\&={\mathcal {L}}\left\{\cos {a\,t}\right\}\,{\mathcal {L}}\left\{\mathrm {e} ^{-2t}\right\}\,{\mathcal {L}}\left\{\mathrm {e} ^{-t}\right\}-{\frac {1}{s+2}}+2\,{\frac {1}{s+1}}\end{aligned}}$

And then take the inverse Laplace transform:

(out of time)